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Denominator of Product_{k=1..n} (2k/(2k+1))^((-1)^A000120(k)).
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%I #33 Aug 30 2015 10:27:13

%S 2,8,28,224,1232,4004,56056,896896,8520512,14910896,328039712,

%T 41004964,1066129064,29851613792,462700013776,1346036403712,

%U 4711127412992,43577928570176,127381637358976,652830891464752,9139632480506528,402143829142287232,9450379984843749952

%N Denominator of Product_{k=1..n} (2k/(2k+1))^((-1)^A000120(k)).

%C For all n, a(n) is even, 2-adic valuation of a(2^n) is 2n+1 and 2-adic valuation of a(3*2^n) is 2.

%H Gheorghe Coserea, <a href="/A261559/b261559.txt">Table of n, a(n) for n = 1..1000</a>

%H Jean-Paul Allouche, <a href="http://algo.inria.fr/seminars/sem92-93/allouche.pdf">Series and infinite products related to binary expansion of integers</a>, December 07, 1992.

%H Jeffrey Shallit, <a href="http://www.cs.uwaterloo.ca/~shallit/Talks/pmc2.ps">Ten Problems I Can't Solve</a>, talk for the University of Waterloo Pure Math Club, July 11 2000.

%t Table[Denominator@ Product[(2 k/(2 k + 1))^((-1)^DigitCount[k, 2, 1]), {k, 1, n}], {n, 23}] (* _Michael De Vlieger_, Aug 26 2015 *)

%o (PARI)

%o n = 22; R(k) = { if (hammingweight(k)%2, (2*k+1)/(2*k), (2*k)/(2*k+1)) };

%o p = vector(n); p[1] = R(1); for(i = 2, #p, p[i] = p[i-1] * R(i));

%o apply(denominator, p)

%Y Cf. A000120, A010060, A094541, A094542, A261505 (numerator).

%K nonn,frac

%O 1,1

%A _Gheorghe Coserea_, Aug 24 2015